I'm attempting a question in my math book (self teaching so don't have a personal tutor to ask). I'm getting confused as to what I'm supposed to be doing. Here's the question: What is the set of values of p for which $p(x^2 + 2) < 2x^2 + 6x + 1$ for all real values of x?

Now I've reworded the question to "find values of p for which $(p - 2)x^2 - 6x + 2p - 1 < 0$ for all real values of x. Because the inequality is $< 0$, I'm guessing that I'm seeking all the parabolas that are below the x-axis. I'm already getting a bit lost as you might tell. Quadratic equations that don't cross the x-axis do not have real roots, so I'm looking at the discriminant $'b^2 - 4ac' < 0$ where $a = (p - 2)$, $b = -6$, and $c = (2p - 1)$. This boils down to the inequality $2p^2 - 5p - 7 > 0$, the solutions of which I compute to be $p < -1$ or $p > \frac{7}{2}$.

The answer in the back of the book is $-1 < p < \frac{7}{2}$ which is not the same as what I have but has the same boundary conditions.

My request is for someone to help me understand each step of the solution please. I'm getting a bit lost and losing the intuition of it all.

The book answer is wrong. If you set p=0, x=-1.5 the inequality is not satisfied. As Robert Israel says, you cannot have p>2 or for large positive $x$ the inequality will fail.
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Ross MillikanApr 21 '11 at 19:54

Thank you for looking at this. I've checked the question, and it's quoted verbatim. Worked through it again slightly differently and getting the same answer (that doesn't match the book). In the unlikely event someone has the errata, the book is Core Maths for Advanced Level by Bostock and Chandler 3rd edition. Exercise 12D, question 3.
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PeteUKApr 21 '11 at 23:44

It's really not uncommon for answers in the back of textbooks to be wrong. In general I think authorns and publishers put much less effort into checking these for correctness than they do in checking the main body of the book.
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Robert IsraelApr 22 '11 at 17:50